Special Heat Transfer Coefficients4.13C Extended Surfaces or Finned Exchangers

Nehal I. Abu-Lail

• How to increase q at fixed T1 and T2?• Newton’s law of cooling states that:

qconvection=hA(T1-T2)• We either increase h (forced convection) or A (using

fins).

Examples include:• Automobile radiators• Ears and nose• Finned heat exchangers

Two common types of fins

Flow

Flow

For a metallic tube with hiand ho, qualitative effect of fins can be seen using:

oometal

iiii AhR

AhR

AU111

• We can neglect R metal as metals have high k

• Assume T to be the same for tube wall and fin

When we have fins, Ao will increase and the outside convective R will be reducedSo what? This can affect our choice of where to place hot and cold fluids in an exchanger arrangement. • Steam inside, hi is high

• Air outside, ho is small• Fins outside, increase Ao

and thus decrease 1/hoAo• q will increase

• Steam outside, ho is high • Air inside, hi is small• Fins outside, increase Ao• Increasing Ao will do little to

increasing q because ho is high already and 1/hoAo is small.

• What we have done so far is just an approximation.• We know that T at the base of fin is different from that at the end of

the fin because of added resistance to heat flow by conduction from the fin tip to the base of the fin.

• Hence, a unit area of fin surface is not efficient as a unit area of bare tube surface at the base of the fin.

• A fin efficiency is derived for various geometries of fins.• We will do it here only for a 1D rectangular fin

Fin Efficiency

• In the limiting case of zero thermal resistance or infinite thermal conductivity ( k ), the temperature of the fin is uniform (maximized) at the base value of Tb

The heat transfer from a fin is maximum and can be expressed as:

Af is the total surface area of the fin.

In reality, the temperature drops along the fin, and thus the heat transfer from the fin is less.

TThAQ bfinfin max,

.

Derivation of fin efficiency Assumptions:Steady state 1 dimensional (x direction),exposed to surrounding at TAt any x, T =TfinThe base of the fin is at To

Let us perform heat balance

cxxx qqq

qc is heat loss by convection

))(( TTxphdxdTkA

dxdTkA xxx

P is perimeter of the fin

Divide by x and take the limit as x goes to zero to get:

),()( TThpdxdT

dxdkA ddTTTTT

kAhp

dxTd

, ),(2

2

0 2

2

kAhp

dxd

0 2

2

kAhp

dxd

To solve this ODE, we need two B.C.s:• At x=0 (base of the fin), T=To, =To-T• If the tip of fin is insulated, then at x=L, d/dx=0 (adiabatic tip, q=0)• If fin losses heat by convection at its tip, that has to equal conduction at the

tip, , note that A will cancel out. This is a little bit involved solution and will not go with it.

• If the fin is infinitely long, T at the tip (x=L) will be = T

)( TThAdxdTkA LL

To solve, this is a second order, homogenous equation

0, 22

2

mdxd 0,) ( 22 mD mD Two real roots, solution will be:

mxmx AA expexp 21 Apply B.C.s, solve for A1 and A2 to get:

mLxLm

TTTT

oo cosh)](cosh[

mkAhp

Call = constant

,exp1

exp1 22

2mx

mLomx

mL

mLo

eee

Remember that:2

coshxx eex

Heat lost by the fin can be expressed as:

nLTThpkAdxdTkAq ox tanh)()( 5.0

0

• For actual fins, T decreases as we approach fin tip• The rate of heat loss per unit area decreases with fin distance

as we approach tip (distance from base increases)• To account for this, we use a fin efficiency (f)• A fin efficiency is the ratio of the actual heat transfer from the

fin to the heat transfer of the fin if the entire fin was at the base Temperature To.

mLmL

TThPLmLTThpkA

o

of

tanh)(tanh)()( 5.0

Lwtk

twhLkAhpmL

21

21

)()22(

In fins which are too thin,

2t<<2w L

kthmL

21

2

• When heat is lost from the tip of the fin (no insulation), we will correct for fin length (Lc)

• Lc=L+(t/2)=Corrected fin length• For longitudinal and circular fins respectively, the charts below give f.

Longitudinal Circular

)(2 xwLA cf ])()[(2 21

21 rrLA cf

Example 4.13-2 Fin efficiency and heat loss from a fin

A circular aluminum fin with k = 222 W/m.K is attached to a copper tube having an OD of 0.04 m. The length of the fin is 0.04 m and its thickness is 2 mm. The OD tube base is at T=523.2 K and the surrounding T of air is 343.2 K. Air has a convective h of 30 W/m2.K. Calculate f and q lost from fin?

Solution• Sketch the problem• Identify variables and assumptions• We will use charts. To do that, we need Lc• Calculate the quantities needed for chart • Read efficiency from chart=0.89• Calculate heat loss

)( TThAq offf

])()[(2 21

21 rrLA cf

Overall Heat Transfer Coefficient for Finned Tubes

• Consider a tube with outside fins• Heat transfer will occur through a series of R• The total heat loss from outside the tube is the

sum of heat loss by convection from base of tube (qt) and heat loss by convection from the fins (qf)

xAq

T1T2

T3

T4hi

ho

kA

ri

ro

)()( 2121 TTAhTTAhqqq ffotoft These can be written in terms of resistances in // as:

RTT

AhAh

TTq

fofto

2121

1

At: Area of tube between finsAf: Area of fins

R here can replace outside convective R (1/hoAo) we have seen before for a finned heat tube exchanger. For Ui, q=UiAi(T1-T4)

Fins

)(1

1

ffto

i

AlmA

iA

i

i

AAhA

AkAx

h

U